This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Consider a Young” s twoslit experiment in which lhe wavelength of light is 33%
smaller than the distance between the slits. {a} How man}.r complete dark things total would be seen on a distant screen?I [Asstune for all these questions that the
screen is as lHJ'E as it needs tn be.) [b] How man}.r total interference rnasima would be seen on a distant semen? (of: Now suppose that the wavelength of light is 134%
larger than the slit separation How man}.r complete darlt hinges would be seen on a
distant staeen'? Solutinn: [at The statement of the problem is equivalent to saying that the wavelength of light is ﬁlii'i of the slit separation. That means that the twoslit destructive interference conditions
bemune dsind=(m+é}i=ﬂ.ﬁﬂ(m+%)d =r si115'=fi.ﬁi(n1+%). Because the value of sintir has to be between —l and +1. onlv certain integers in will
result in equations that can be solved for an actual angle. Thwe allowable values of In
tell us how ntan];r tiarlt frir : s there are. For the deal: fringes. values of m. = fl.J_l.‘I all warla sn there are four dark fringes. {b} This time we use the constructive interference condition dsinl'i' _ innit _ ﬂiﬂnni _: sinl'i' _ littlﬂm. Again Tinting that aim? has tn lie. between :l:. we find that. the nlluwaljle, integers are
in = [Li], which means that there are three bright fringes visible on the screen. {e} In this part we‘re told that the wavelength of light is 134% larger than the slit separa
tion: in other words. it = 2.34s. Then the destructive interference condition is I l l
nisinli' = (m + i) a = ass [m + E) s =:~ sinti = assm + ii“ This time. there is no integer in that will produce a solvable equation. So there are zero visible dark hinges. In other words. the central max is a huge smear across the HL'TE'E'TI. ...
View
Full Document
 Spring '08
 SEATON

Click to edit the document details